Motion of two bodies at both ends of a string

By conservation of momentum the centre of mass of the system must move to the right with constant speed

$displaystyle frac {m_b}{m_a+m_b}v_0 = frac {m_b}{M}v_0$

where $M=m_a+m_b$.

If you work in the reference frame in which the COM is stationary (note that this is an inertial reference frame) then you will find that $m_a$, which is at a distance $frac {m_b} M l$ from the COM, initially moves to left with speed $frac {m_b} M v_0$, so has angular speed $frac {v_0} l$ rad/s anticlockwise about the COM.

Similarly $m_b$, which is at distance $frac {m_a} M l$ from the COM, initially moves to right with speed $frac {m_a} M v_0$, so it also has angular speed $frac {v_0} l$ rad/s anticlockwise about the COM.

So both masses revolve about the COM with the same constant angular speed $frac {v_0} l$ rad/s. Once you know this you can calculate the tension in the string - and also convince yourself that the string does not collapse.

First, consider the initial motion, and try to imagine the system as a rigid body

The center of mass must have a horizontal velocity equal to

$$ v_{rm COM} = frac{a}{ell} v_0 $$

The velocity is maintained through-out the motion since no external forces act here.

In addition, the motion is decomposed to a rotation about the center of mass with a rotational speed

$$ omega = frac{v_0}{ell} $$

This means the motion of mass $m_a$ tracks a circle around the center of mass with radius $a$ and the mass $m_b$ tracks a circle around the center of mass with radius $b$.

The tension is such as to force both of these motions

$$ T = m_a omega^2 a = m_b omega^2 b $$

and remember that the center of mass is defined by $a = frac{m_b}{m_a+m_b} ell$ and $b = frac{m_a}{m_a+m_b} ell$.